The slope of the tangent to the curve `y=x^(3) -x +1` at the point whose x-coordinate is 2 is

Options

A. `-11`
B. `(1)/(11)`
C. 11
D. `3x^(2)-1`

Answer

To find the slope of the tangent to the curve \( y = x^3 - x + 1 \) at the point where the x-coordinate is 2, we need to follow these steps:

### Step 1: Differentiate the function
We start by finding the derivative of the function \( y \) with respect to \( x \). The derivative \( \frac{dy}{dx} \) gives us the slope of the tangent line at any point on the curve.

Given:
\[ y = x^3 - x + 1 \]

We differentiate:
\[ \frac{dy}{dx} = \frac{d}{dx}(x^3) - \frac{d}{dx}(x) + \frac{d}{dx}(1) \]

Calculating each term:
- The derivative of \( x^3 \) is \( 3x^2 \).
- The derivative of \( -x \) is \( -1 \).
- The derivative of \( 1 \) is \( 0 \).

So, we have:
\[ \frac{dy}{dx} = 3x^2 - 1 \]

### Step 2: Substitute \( x = 2 \)
Now, we need to find the slope of the tangent at the point where \( x = 2 \). We substitute \( x = 2 \) into the derivative we found.

\[ \frac{dy}{dx} \bigg|_{x=2} = 3(2^2) - 1 \]

Calculating \( 2^2 \):
\[ 2^2 = 4 \]

Now substituting back:
\[ \frac{dy}{dx} \bigg|_{x=2} = 3(4) - 1 = 12 - 1 = 11 \]

### Conclusion
The slope of the tangent to the curve at the point where the x-coordinate is 2 is:
\[ \boxed{11} \]